1
$\begingroup$

Let $\mathfrak{g}$ be a nilpotent Lie algebra. It is possible to find the Lie algebra of derivations of $\mathfrak{g}$ denoted $Der\mathfrak{g}.$

Then we could consider the maximal abelian subalgebra of the Lie algebra $Der\mathfrak{g}$ consisting of semisimple elements. It turned out to be that such abelian Lie algebra is called the maximal torus of derivations. Besides the dimension of the maximal torus is known to be the rank of the nilpotent Lie algebra $\mathfrak{g}$.

Could you please tell me if I could use the rank of a particular nilpotent Lie algebra to figure out the maximal possible dimension of the corresponding solvable Lie algebra having the nilpotent Lie algebra as the nilradical?

$\endgroup$
0
$\begingroup$

For safety I assume we are over a field of characteristic zero; I expect you assume this and that the Lie algebras are finite-dimensional.

If $\mathfrak{g}$ is nilpotent and has rank $d$, and if $\mathfrak{h}$ is solvable with nilpotent radical $\mathfrak{g}$, then $\dim(\mathfrak{h})\le d+\dim(\mathfrak{g})$. You get equality just by picking $\mathfrak{h}=\mathfrak{d}\ltimes\mathfrak{g}$ where $\mathfrak{d}$ is a $d$-dimensional abelian subalgebra of $\mathrm{Der}(\mathfrak{g})$ consisting of semisimple endomorphisms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.